Verifying that $ prod_{j=1}^{infty} frac{1}{1-q^j} = prod_{j=1}^{infty} frac{1}{(1-q^{2j-1})(1-q^{2j})}$
$begingroup$
On page 165 of Chapter 13, how was the equality made from line 1 to line 2?
https://archive.org/details/NumberTheory_862/page/n173
Namely, how $$ prod_{j=1}^{infty} frac{1}{1-q^j} = prod_{j=1}^{infty} frac{1}{(1-q^{2j-1})(1-q^{2j})}$$
infinite-product integer-partitions
edited Dec 17 '18 at 0:42
Blue
48.4k870154
asked Dec 17 '18 at 0:25
zodrosszodross
1546
$endgroup$
add a comment |
$begingroup$
On page 165 of Chapter 13, how was the equality made from line 1 to line 2?
https://archive.org/details/NumberTheory_862/page/n173
Namely, how $$ prod_{j=1}^{infty} frac{1}{1-q^j} = prod_{j=1}^{infty} frac{1}{(1-q^{2j-1})(1-q^{2j})}$$
infinite-product integer-partitions
edited Dec 17 '18 at 0:42
Blue
48.4k870154
asked Dec 17 '18 at 0:25
zodrosszodross
1546
$endgroup$
add a comment |
$begingroup$
On page 165 of Chapter 13, how was the equality made from line 1 to line 2?
https://archive.org/details/NumberTheory_862/page/n173
Namely, how $$ prod_{j=1}^{infty} frac{1}{1-q^j} = prod_{j=1}^{infty} frac{1}{(1-q^{2j-1})(1-q^{2j})}$$
infinite-product integer-partitions
edited Dec 17 '18 at 0:42
Blue
48.4k870154
asked Dec 17 '18 at 0:25
zodrosszodross
1546
$endgroup$
On page 165 of Chapter 13, how was the equality made from line 1 to line 2?
https://archive.org/details/NumberTheory_862/page/n173
Namely, how $$ prod_{j=1}^{infty} frac{1}{1-q^j} = prod_{j=1}^{infty} frac{1}{(1-q^{2j-1})(1-q^{2j})}$$
infinite-product integer-partitions
infinite-product integer-partitions
edited Dec 17 '18 at 0:42
Blue
48.4k870154
asked Dec 17 '18 at 0:25
zodrosszodross
1546
edited Dec 17 '18 at 0:42
Blue
48.4k870154
asked Dec 17 '18 at 0:25
zodrosszodross
1546
edited Dec 17 '18 at 0:42
Blue
48.4k870154
edited Dec 17 '18 at 0:42
Blue
48.4k870154
edited Dec 17 '18 at 0:42
Blue
48.4k870154
48.4k870154
asked Dec 17 '18 at 0:25
zodrosszodross
1546
asked Dec 17 '18 at 0:25
zodrosszodross
1546
asked Dec 17 '18 at 0:25
zodrosszodross
1546
1546
add a comment |
add a comment |
1 Answer
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$begingroup$
Simply pair-off factors and use the fact that multiplication is commutative:
begin{align*}
prod_{j=1}^{infty}frac{1}{(1-q^{j})} & = frac{1}{1-q}frac{1}{1-q^2}frac{1}{1-q^3}frac{1}{1-q^4}cdots\
& = left(frac{1}{1-q}frac{1}{1-q^2}right)left(frac{1}{1-q^3}frac{1}{1-q^4}right)cdots\
& = left(frac{1}{1-q^{2}}frac{1}{1-q}right)left(frac{1}{1-q^4}frac{1}{1-q^{3}}right)cdots\
& = prod_{j=1}^{infty}frac{1}{(1-q^{2j})(1-q^{2j-1})}.
end{align*}
answered Dec 17 '18 at 0:30
Will RWill R
6,59731429
$endgroup$
$begingroup$
Oh, I see. Thank you.
$endgroup$
– zodross
Dec 17 '18 at 0:32
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Simply pair-off factors and use the fact that multiplication is commutative:
begin{align*}
prod_{j=1}^{infty}frac{1}{(1-q^{j})} & = frac{1}{1-q}frac{1}{1-q^2}frac{1}{1-q^3}frac{1}{1-q^4}cdots\
& = left(frac{1}{1-q}frac{1}{1-q^2}right)left(frac{1}{1-q^3}frac{1}{1-q^4}right)cdots\
& = left(frac{1}{1-q^{2}}frac{1}{1-q}right)left(frac{1}{1-q^4}frac{1}{1-q^{3}}right)cdots\
& = prod_{j=1}^{infty}frac{1}{(1-q^{2j})(1-q^{2j-1})}.
end{align*}
answered Dec 17 '18 at 0:30
Will RWill R
6,59731429
$endgroup$
$begingroup$
Oh, I see. Thank you.
$endgroup$
– zodross
Dec 17 '18 at 0:32
add a comment |
$begingroup$
Simply pair-off factors and use the fact that multiplication is commutative:
begin{align*}
prod_{j=1}^{infty}frac{1}{(1-q^{j})} & = frac{1}{1-q}frac{1}{1-q^2}frac{1}{1-q^3}frac{1}{1-q^4}cdots\
& = left(frac{1}{1-q}frac{1}{1-q^2}right)left(frac{1}{1-q^3}frac{1}{1-q^4}right)cdots\
& = left(frac{1}{1-q^{2}}frac{1}{1-q}right)left(frac{1}{1-q^4}frac{1}{1-q^{3}}right)cdots\
& = prod_{j=1}^{infty}frac{1}{(1-q^{2j})(1-q^{2j-1})}.
end{align*}
answered Dec 17 '18 at 0:30
Will RWill R
6,59731429
$endgroup$
$begingroup$
Oh, I see. Thank you.
$endgroup$
– zodross
Dec 17 '18 at 0:32
add a comment |
$begingroup$
Simply pair-off factors and use the fact that multiplication is commutative:
begin{align*}
prod_{j=1}^{infty}frac{1}{(1-q^{j})} & = frac{1}{1-q}frac{1}{1-q^2}frac{1}{1-q^3}frac{1}{1-q^4}cdots\
& = left(frac{1}{1-q}frac{1}{1-q^2}right)left(frac{1}{1-q^3}frac{1}{1-q^4}right)cdots\
& = left(frac{1}{1-q^{2}}frac{1}{1-q}right)left(frac{1}{1-q^4}frac{1}{1-q^{3}}right)cdots\
& = prod_{j=1}^{infty}frac{1}{(1-q^{2j})(1-q^{2j-1})}.
end{align*}
answered Dec 17 '18 at 0:30
Will RWill R
6,59731429
$endgroup$
Simply pair-off factors and use the fact that multiplication is commutative:
begin{align*}
prod_{j=1}^{infty}frac{1}{(1-q^{j})} & = frac{1}{1-q}frac{1}{1-q^2}frac{1}{1-q^3}frac{1}{1-q^4}cdots\
& = left(frac{1}{1-q}frac{1}{1-q^2}right)left(frac{1}{1-q^3}frac{1}{1-q^4}right)cdots\
& = left(frac{1}{1-q^{2}}frac{1}{1-q}right)left(frac{1}{1-q^4}frac{1}{1-q^{3}}right)cdots\
& = prod_{j=1}^{infty}frac{1}{(1-q^{2j})(1-q^{2j-1})}.
end{align*}
answered Dec 17 '18 at 0:30
Will RWill R
6,59731429
answered Dec 17 '18 at 0:30
Will RWill R
6,59731429
answered Dec 17 '18 at 0:30
Will RWill R
6,59731429
answered Dec 17 '18 at 0:30
Will RWill R
6,59731429
6,59731429
$begingroup$
Oh, I see. Thank you.
$endgroup$
– zodross
Dec 17 '18 at 0:32
add a comment |
$begingroup$
Oh, I see. Thank you.
$endgroup$
– zodross
Dec 17 '18 at 0:32
$begingroup$
Oh, I see. Thank you.
$endgroup$
– zodross
Dec 17 '18 at 0:32
$begingroup$
Oh, I see. Thank you.
$endgroup$
– zodross
Dec 17 '18 at 0:32
add a comment |
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